If θ ∈[-2 π, 2 π], then the number of solutions of 2 √2 cos ² θ+(2-√6) cos θ-√3=0, is equal to:

Solution

Maths sir · 3 steps

Treat cos θ as x, factor the quadratic, then count angles for each root over two full turns.

1. Factor the quadratic. Put x = cos θ. The middle term (2-√6)x splits as 2x - √6x, and the equation factors neatly.

   2√2x² + (2-√6)x - √3 = (√2x + 1)(2x - √3) = 0

2. Both roots allowed. So cos θ = √3/2 or cos θ = -1/√2. Both lie strictly between -1 and 1. Each one gives 2 angles per turn, so 4 angles in [-2π, 2π].

   Value of cos θ	Solutions in [-2π, 2π]	Count
   √3/2	±π/6, ±11π/6	4
   -1/√2	±3π/4, ±5π/4	4

3. So. The total number of solutions is 4 + 4 = 8, option (a).

The idea to keep. For a quadratic in cos θ, first keep only the roots in [-1, 1]. Each value strictly inside gives 2 angles per 2π, so count the turns in the interval and multiply.

Answer from the NTA final answer key. Solution by PrepWiser.